Chebyshev expansion of $\log(1 + x)$

Question

I was reading a Wikipedia article on Chebyshev polynomials and got stuck in around the end of the article where the author takes advantage of orthogonality to compute the coefficients of the Chebyshev expansion of $ \log (x+1) $. I will be happy if someone explains the process to get $$a_0 = - \log(2)\quad\mbox{and}\quad a_n = \frac{- \pi (-1)^n}{n} \quad\mbox{ for $ n \in \mathbb{N} $.}$$ Any help is appreciated.

Answer 1

Note that for $t\in (0,\pi)$, $$ \begin{align} \log(1+\cos(t)) &=\log\left(\frac{e^{it}+2+e^{-it}}{2}\right)\\ &=\log\left(1+e^{it}\right)+\log\left(1+e^{-it}\right)-\log(2)\\ &=\sum_{n=1}^\infty(-1)^{n+1}\frac{e^{int}}{n}+\sum_{n=1}^\infty(-1)^{n+1}\frac{e^{-int}}{n}-\log(2)\\ &=2\sum_{n=1}^\infty(-1)^{n+1}\frac{\cos(nt)}{n}-\log(2). \end{align} $$ Hence, by letting $x=\cos(t)\in (-1,1)$, we obtain $$\log(1+x)=\sum_{n=1}^\infty\frac{2(-1)^{n+1}}{n}T_n(x)-\log(2)T_0(x)$$ which implies that $$a_0=-\log(2)\quad\mbox{and}\quad a_n=\frac{2(-1)^{n+1}}{n}\mbox{ for $n>0$}$$ where $\ln(1+x)=\sum_{n\geq 0}a_n T_n(x)$, which is a bit different from what we read on the wikipage.

P.S. Going back to the wikipage: \begin{align*} \int_{-1}^1\frac{\log(1+x)T_m(x)}{\sqrt{1-x^2}}\,dx&=\int_{\pi}^0\frac{\log(1+\cos(t))\cos(mt)}{|\sin(t)|}\,(-\sin(t)dt)\\ &=\int_0^{\pi}\left(2\sum_{n=1}^\infty(-1)^{n+1}\frac{\cos(nt)}{n}-\log(2)\right)\cos(mt) dt\\ &=\begin{cases} -\pi\log(2) &\mbox{if $m=0$,}\\ \dfrac{(-1)^{m+1}\pi}{m}&\mbox{if $m>0$.} \end{cases} \end{align*} On the other hand if $\ln(1+x)=\sum_{n\geq 0}a_n T_n(x)$, \begin{align*} \int_{-1}^1\frac{\log(1+x)T_m(x)}{\sqrt{1-x^2}}\,dx&=a_m\int_{-1}^1\frac{T_m(x)^2}{\sqrt{1-x^2}}\,dx =\begin{cases} \pi a_0 &\mbox{if $m=0$,}\\ \dfrac{\pi}{2}a_m&\mbox{if $m>0$.} \end{cases} \end{align*}